Question

at 25 oc the solubility of calcium carbonate is 9.33 x 10-5 mol/l. calculate the value of ksp at this temperature. give your answer in scientific notation to 2 significant figures (even though this is strictly incorrect).

Answer 1

Explanation:The solubility product constant (Ksp) for calcium carbonate (CaCO3) is given by the expression:CaCO3(s) ⇌ Ca2+(aq) + CO32-(aq)

The solubility of CaCO3 is 9.33 x 10^-5 mol/L at 25°C. Therefore, the concentration of Ca2+ and CO32- ions in the saturated solution is also 9.33 x 10^-5 mol/L.

Using the equation for Ksp, we have:

Ksp = [Ca2+][CO32-]

Ksp = (9.33 x 10^-5 mol/L)(9.33 x 10^-5 mol/L)

Ksp = 8.68 x 10^-10

Therefore, the value of Ksp for calcium carbonate at 25°C is 8.68 x 10^-10, rounded to 2 significant figures. I hope this helps :)